Question

三角形 $ABC$ 在 $B$ 处为直角。点 $D$ 是从 $B$ 向斜边作高的垂足，且 $AD = 3$ 、 $DC = 4$ 。 $\triangle ABC$ 的面积是多少？

Triangle $ABC$ has a right angle at $B.$ Point $D$ is the foot of the altitude from $B,$ $AD = 3,$ and $DC = 4.$ What is the area of $\triangle ABC?$

$4\sqrt{3}$

$7\sqrt{3}$

$21$

$14\sqrt{3}$

$42$

对直角三角形中从直角顶点到斜边的高，有 $BD^2 = AD \cdot DC = 3 \cdot 4 = 12,$ 所以 $BD = 2\sqrt{3}$ 。

斜边满足 $AC = 3 + 4 = 7$ ，所以面积为 $\dfrac12 \cdot 7 \cdot 2\sqrt3 = 7\sqrt3.$

所以正确答案是 B。

For the altitude from the right angle to the hypotenuse, $BD^2 = AD \cdot DC = 3 \cdot 4 = 12,$ so $BD = 2\sqrt{3}.$

The hypotenuse is $AC = 3 + 4 = 7,$ so the area is $\dfrac12 \cdot 7 \cdot 2\sqrt3 = 7\sqrt3.$

Thus, the correct answer is B.